Trial Questions (1 Viewer)

[scardizzle]

Hey guys,

Im having trouble with these 2 problems from my school's past paper (no solutions "-_- )

1. Differentiate y= ln(tan3x) epressing your final answer in terms of cosec3x. I can do the differentiation part but i end up with 3/(cos3xsin3x) = 6/sin6x ?

2. the polynomial P(x) is given by P(x) = x^3 - x^2 - 8x + 12
i) Factorise P(x) completely given P(x) = 0 has a repeated root
ii) The polynomial Q(x) has the form Q(x) = P(x)(x+a) with P(x) as given above and where the constant a is chosen so the Q(x) >/= 0 for all real values of x. Find all possible values of a.

I can do part i) but no idea where to start from part ii)

Thanks in advance

 

[Trebla]

Hey guys,

Im having trouble with these 2 problems from my school's past paper (no solutions "-_- )

1. Differentiate y= ln(tan3x) epressing your final answer in terms of cosec3x. I can do the differentiation part but i end up with 3/(cos3xsin3x) = 6/sin6x ?

2. the polynomial P(x) is given by P(x) = x^3 - x^2 - 8x + 12
i) Factorise P(x) completely given P(x) = 0 has a repeated root
ii) The polynomial Q(x) has the form Q(x) = P(x)(x+a) with P(x) as given above and where the constant a is chosen so the Q(x) >/= 0 for all real values of x. Find all possible values of x.

I can do part i) but no idea where to start from part ii)

Thanks in advance

For 1) I think it should say "cosec 6x"

2) i) Not sure if this is in the Ext 1 course but for a repeated root P'(x) = P(x) = 0 is satisfied.
This gives x = 2 as a root and then use the sum of roots to find the third root.

ii) Not sure if I interpreted this correctly, (I think you mean all possible values of a?)
Q(x) = (x + a)(x³ - x² - 8x + 12) ≥ 0
=> (x + a)(x - 2)²(x + 3) ≥ 0
For expression to hold true we require (x + a)(x + 3) ≥ 0
The only way this expression holds true for all real x is when a = 3 (which gives a perfect square which is always non-negative)

 

[cutemouse]

Not sure if this is in the Ext 1 course but for a repeated root P'(x) = P(x) = 0 is satisfied.

It is.

 

[gurmies]

Not sure about that...I thought multiplicity was an extension II concept...

 

[cutemouse]

Not sure about that...I thought multiplicity was an extension II concept...

I think in Ext 1 you only consider roots of multiplicity of 2. ie. If a is a double root of P(x) then P(a)=P'(a)=0. But in Ext 2 you can be asked roots of other multiplicity.

This was in our Year 11 preliminary exam for Ext 1. The question was something like, if a is a double root of P(x), then prove that P(a)=P'(a)=0.

As was finding a^3+b^3+c^3, given that a, b and c are roots of P(x)=...

 

[Aquawhite]

I think in Ext 1 you only consider roots of multiplicity of 2. ie. If a is a double root of P(x) then P(a)=P'(a)=0. But in Ext 2 you can be asked roots of other multiplicity.

This was in our Year 11 preliminary exam for Ext 1. The question was something like, if a is a double root of P(x), then prove that P(a)=P'(a)=0.

As was finding a^3+b^3+c^3, given that a, b and c are roots of P(x)=...

In Ext. 1 we are often asked to do that, yes... but I wouldn't name it as multiplicy though...

 

[addikaye03]

I think in Ext 1 you only consider roots of multiplicity of 2. ie. If a is a double root of P(x) then P(a)=P'(a)=0. But in Ext 2 you can be asked roots of other multiplicity.

This was in our Year 11 preliminary exam for Ext 1. The question was something like, if a is a double root of P(x), then prove that P(a)=P'(a)=0.

As was finding a^3+b^3+c^3, given that a, b and c are roots of P(x)=...

Theres also a method called "Newtons Sums" to find the sum of roots when given a P(x)

 

[Aquawhite]

Theres also a method called "Newtons Sums" to find the sum of roots when given a P(x)

What's that/how's it work?

I only know the regular sum of roots, sum of roots two at a time (etc...), product of roots. blah blah blah.

 

[addikaye03]

What's that/how's it work?

I only know the regular sum of roots, sum of roots two at a time (etc...), product of roots. blah blah blah.

It's more used for Q where:
-P(x) given
-trying to find x1,x2,x3,...,xn

Link: Newton's sums - AoPSWiki

 

[Michaelmoo]

1) Unless anyone else can think of another way, I think the question should've been express it in terms of cosec6x. In that case Uve got the answer:

6cosec(6x)

2) (i)The way they teach it in 3 unit is as follows:

• Look at the degree, it is to a power of 3 so there are three roots
• It says there is a repeated root, now this has root has to appear either twice or 3 times
• You know that the three roots multipled should give the constant term (i.e. 12). However, no number when cubed gives twelve, so there can't be triple root.
• So now we know that there is a double root and a single root. Call the double root "a" and the single root "b".
• Call the roots a,a and b

Lets do a sum of the roots:

a + a + b = 1
2a + b = 1
b = 1 - 2a -------------------- (1)

Sum of the roots 2 at a time:

a^2 + ab + ab = -8
a^2 + 2ab = -8 ----------------- (2)

Now (1) into (2)

a^2 + 2a(1 - 2a) = -8
a^2 + 2a - 4a^2 = -8
3a^2 - 2a - 8 = 0

Solve this quadratic you get a = 2 (thus b = -3) or a = -4/3 (and b = 11/3)

However when you consider the multiplication of roots..

a x a x b = -12
(a^2)b = -12

This equation is only satisfied for the a =2 b =-3 combination. Thus these are the roots:

Hence factorising gives (x+3)(x-2)^2

The derivative/root method is only 4 unit I think..

(ii) Draw the graph above, you notice it starts to go below the x- axis at x = -3. So we need to make that root a double root (so it turns back up and doesn't go below the x-axis, i.e. remains greater than 0). i.e. multiple the equation by (x+3) , i.e. a = 3